Sequences in their rigorous form, plus the Cauchy criterion — convergence detected without knowing the limit. This is the deep idea the analysis stream adds beyond ordinary calculus, and it is what makes $\mathbb{R}$ different from $\mathbb{Q}$.
Sometimes you must know a sequence converges without naming its limit — the limit may not have a closed form at all. The Cauchy criterion does exactly that, and it is how the completeness of $\mathbb{R}$ is most usefully expressed and used in practice.
Every iterative algorithm is a sequence — Newton's method, fixed-point iteration, gradient descent, machine-learning training loops. "Cauchy" is the practical test that successive iterates stop changing, which is exactly the stopping criterion these algorithms use.
Cantor and Dedekind used Cauchy sequences (and Dedekind cuts, respectively) to construct $\mathbb{R}$ rigorously from $\mathbb{Q}$ in the 1870s. Zeno's 450 BC paradox of Achilles and the tortoise is resolved by the limit of exactly such a sequence of positions.
$\lim a_n=L$ means: $\forall\varepsilon>0\ \exists N\in\mathbb{N}$ such that $n\ge N\Rightarrow|a_n-L|<\varepsilon$. Informally, the tail of the sequence eventually stays as close to $L$ as you like.
If $a_n\to L$ and $b_n\to M$, then $a_n+b_n\to L+M$, $a_nb_n\to LM$, and (if $M\ne0$) $a_n/b_n\to L/M$. Every one of these follows from the triangle inequality applied to the ε–N definitions of the two sequences — the same "add and subtract" trick from Unit 1's absolute-value work.
Prove $\lim\dfrac{2n+1}{n}=2$.
An increasing sequence bounded above converges, to $\sup\{a_n\}$ (symmetrically for decreasing and bounded below). This is really the completeness axiom of $\mathbb{R}$ restated for sequences: the supremum exists by completeness, and the tail must get close to it or else a smaller upper bound would exist.
Every bounded sequence of real numbers has a convergent subsequence. (Proof idea: repeatedly bisect an interval containing infinitely many terms, always keeping the half that still contains infinitely many terms; the nested intervals pin down a limit.)
$(a_n)$ is Cauchy if $\forall\varepsilon>0\ \exists N$ such that $m,n\ge N\Rightarrow|a_m-a_n|<\varepsilon$ — terms compare to each other, with no reference to a limit.
A sequence of real numbers converges if and only if it is Cauchy.
• "Cauchy" compares terms to each other, not to a (possibly unknown)
limit — do not write $|a_n-L|$ in a Cauchy proof.
• Cauchy implies convergent only because $\mathbb{R}$ is complete — the sequence
of rational truncations of $\sqrt2$ is Cauchy in $\mathbb{Q}$ but has no limit
in $\mathbb{Q}$.
• A bounded sequence need not converge unless it is also monotone, or you only
extract a convergent subsequence (Bolzano–Weierstrass), not the whole
sequence.
• In a limit-law computation, each individual limit must be known to exist
first — you cannot split $\lim(a_n/b_n)$ this way if $b_n\to0$.
Prove $\lim\dfrac{n}{n+1}=1$.
Show $a_1=\sqrt2,\ a_{n+1}=\sqrt{2+a_n}$ converges and find the limit.
Why is $a_n=(-1)^n$ not Cauchy?
Given $a_n\to3$ and $b_n\to-2$, find $\lim\dfrac{2a_n+b_n^2}{a_n b_n}$.
$a_n=\sin(n)$. Explain why a convergent subsequence must exist, without finding one explicitly.
If $|a_{n+1}-a_n|\le\dfrac{1}{2^n}$ for all $n$, show $(a_n)$ is Cauchy.
Let $a_n$ be the decimal truncation of $\sqrt2$ to $n$ places (so $a_1=1.4,\ a_2=1.41,\dots$). Explain why $(a_n)$ is Cauchy in $\mathbb{Q}$ but the completeness theorem "converges iff Cauchy" does not hold in $\mathbb{Q}$.
Prove: if every subsequence of $(a_n)$ has a further subsequence converging to the same limit $L$, then $a_n\to L$.
Course coverage. This unit is part of the complete MAT159H1 topic map. It should cover the following items before you move on:
Use the current UofT MAT159H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.
The sequence spine is: convergence implies Cauchy; completeness gives Cauchy implies convergence; monotone bounded sequences are the most common existence tool.
Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.
Show $a_n=1/n$ is Cauchy.
Explain why $a_n=(-1)^n$ is not Cauchy.
If two subsequences have different limits, what follows about the original sequence?
For $a_{n+1}=(a_n+2/a_n)/2$ with $a_1>0$, what would you try to prove before taking limits?